Consider a galvanic cell consisting of the following two redoxcouples: Ag + (0.0

Question

Consider a galvanic cell consisting of the following two redoxcouples:
Ag+ (0.010M) + e- -->Ag         Eo = +0.80 V
Bi3+ (0.0010M) + 3e- -->Bi      Eo = +0.20 V

a. Write the equation for the half-reaction occurring at thecathode.
b. Write the equation for the half-reaction occurring at theanode.
c. Write the equation for the cell reaction.
d. What is the standard cell potential,Eocell, for the cell?
e. Realizing the nonstandard concentrations, what is theactual cell potential, Ecell, for the cell?

Explanation / Answer

The word galvanic cell is a big clue in this problem, as it tellsyou that the reaction will be spontaneous, and therefore the cellpotential will be positive. Ag+ + e- -->Ag         Eo = +0.80 V Bi3+ + 3e- -->Bi      Eo = +0.20 V The way these are written, both would be the reaction that occursat the cathode, because reduction is occurring. For a cell, it isnecessary to have one reduction reaction and one oxidationreaction, so one of these reactions needs to be turned around. Hereis where you use the piece of information obtained from thedirections, that is, that the cell potential will be positive.Looking at the potentials you have, which one can you turn aroundand when you add them, the outcome will be positive? You have to turn around the Bi reaction because it is less positivethan the Ag reaction. So your new reactions are: Ag+ + e- -->Ag         Eo = +0.80 V Bi --> Bi3+ +3e-      Eo= -0.20 V When you turn the reaction around, you change the sign on the cellpotential. A) Knowing that reduction occurs at the cathode, and reduction isthe gain of electrons, you can tell that the Ag reaction to formAg+ occurs at the cathode. B) That leaves the Bi reaction to occur at the anode. C) To write the cell reaction, you balance electrons for the twoequations. In this case, you have 3 electrons in the Bi reactionand 1 in the Ag reaction, so you're going to need to multiply theAg reaction by 3 so that the electrons are the same: 3(Ag+ + e- --> Ag) Bi --> Bi3+ +3e-    Then you add the two reactions together: 3Ag+ + 3e- + Bi -->3Ag + Bi3+ + 3e- Then cancel electrons to get your overall reation: 3Ag+ + Bi --> 3Ag + Bi3+ D) The standard cell potential for this cell is the two cellpotentials added, so: Ag+ + e- -->Ag         Eo = +0.80 V Bi --> Bi3+ +3e-      Eo= -0.20V                              +0.80 + (-0.20) = +0.60 V E) Because the reaction occurs with nonstandard molarities, it isnecessary to account for this using: E=Eo - (RT/nF)lnQ E=new E Eo=standard E R=8.314 T=temperature n=moles of electrons/reaction F=96500 C Q=[products (not solid or liquid)]/[reactants] (These are raised tothe power of their coefficients.) Since it does not specify nonstandard conditions other thanmolarity, you can assume that it's at 25 C. This means that you canused a condensed version of that formula: E=Eo-(0.0592/n)logQ You have Eo, and found n earlier in the problem. AndQ=[Bi3+]/[Ag+]3 The problem gave that: [Ag+] =0.010M [Bi3+] =0.0010M So, Q= [0.0010]/[0.010]3 Eo=+0.60 n=3 E=+0.60-(0.0592/3)log(0.0010/0.0103)

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